3.390 \(\int \csc ^3(e+f x) (b \sec (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=113 \[ -\frac{5 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{4 f}-\frac{5 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{4 f}+\frac{5 b \sqrt{b \sec (e+f x)}}{2 f}-\frac{\cot ^2(e+f x) (b \sec (e+f x))^{5/2}}{2 b f} \]

[Out]

(-5*b^(3/2)*ArcTan[Sqrt[b*Sec[e + f*x]]/Sqrt[b]])/(4*f) - (5*b^(3/2)*ArcTanh[Sqrt[b*Sec[e + f*x]]/Sqrt[b]])/(4
*f) + (5*b*Sqrt[b*Sec[e + f*x]])/(2*f) - (Cot[e + f*x]^2*(b*Sec[e + f*x])^(5/2))/(2*b*f)

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Rubi [A]  time = 0.0839355, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2622, 288, 321, 329, 212, 206, 203} \[ -\frac{5 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{4 f}-\frac{5 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{4 f}+\frac{5 b \sqrt{b \sec (e+f x)}}{2 f}-\frac{\cot ^2(e+f x) (b \sec (e+f x))^{5/2}}{2 b f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^3*(b*Sec[e + f*x])^(3/2),x]

[Out]

(-5*b^(3/2)*ArcTan[Sqrt[b*Sec[e + f*x]]/Sqrt[b]])/(4*f) - (5*b^(3/2)*ArcTanh[Sqrt[b*Sec[e + f*x]]/Sqrt[b]])/(4
*f) + (5*b*Sqrt[b*Sec[e + f*x]])/(2*f) - (Cot[e + f*x]^2*(b*Sec[e + f*x])^(5/2))/(2*b*f)

Rule 2622

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \csc ^3(e+f x) (b \sec (e+f x))^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^{7/2}}{\left (-1+\frac{x^2}{b^2}\right )^2} \, dx,x,b \sec (e+f x)\right )}{b^3 f}\\ &=-\frac{\cot ^2(e+f x) (b \sec (e+f x))^{5/2}}{2 b f}+\frac{5 \operatorname{Subst}\left (\int \frac{x^{3/2}}{-1+\frac{x^2}{b^2}} \, dx,x,b \sec (e+f x)\right )}{4 b f}\\ &=\frac{5 b \sqrt{b \sec (e+f x)}}{2 f}-\frac{\cot ^2(e+f x) (b \sec (e+f x))^{5/2}}{2 b f}+\frac{(5 b) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (-1+\frac{x^2}{b^2}\right )} \, dx,x,b \sec (e+f x)\right )}{4 f}\\ &=\frac{5 b \sqrt{b \sec (e+f x)}}{2 f}-\frac{\cot ^2(e+f x) (b \sec (e+f x))^{5/2}}{2 b f}+\frac{(5 b) \operatorname{Subst}\left (\int \frac{1}{-1+\frac{x^4}{b^2}} \, dx,x,\sqrt{b \sec (e+f x)}\right )}{2 f}\\ &=\frac{5 b \sqrt{b \sec (e+f x)}}{2 f}-\frac{\cot ^2(e+f x) (b \sec (e+f x))^{5/2}}{2 b f}-\frac{\left (5 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{b-x^2} \, dx,x,\sqrt{b \sec (e+f x)}\right )}{4 f}-\frac{\left (5 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{b+x^2} \, dx,x,\sqrt{b \sec (e+f x)}\right )}{4 f}\\ &=-\frac{5 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{4 f}-\frac{5 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{4 f}+\frac{5 b \sqrt{b \sec (e+f x)}}{2 f}-\frac{\cot ^2(e+f x) (b \sec (e+f x))^{5/2}}{2 b f}\\ \end{align*}

Mathematica [A]  time = 2.27884, size = 97, normalized size = 0.86 \[ -\frac{(b \sec (e+f x))^{3/2} \left (-5 \log \left (1-\sqrt{\sec (e+f x)}\right )+5 \log \left (\sqrt{\sec (e+f x)}+1\right )+4 \left (\csc ^2(e+f x)-5\right ) \sqrt{\sec (e+f x)}+10 \tan ^{-1}\left (\sqrt{\sec (e+f x)}\right )\right )}{8 f \sec ^{\frac{3}{2}}(e+f x)} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^3*(b*Sec[e + f*x])^(3/2),x]

[Out]

-((10*ArcTan[Sqrt[Sec[e + f*x]]] - 5*Log[1 - Sqrt[Sec[e + f*x]]] + 5*Log[1 + Sqrt[Sec[e + f*x]]] + 4*(-5 + Csc
[e + f*x]^2)*Sqrt[Sec[e + f*x]])*(b*Sec[e + f*x])^(3/2))/(8*f*Sec[e + f*x]^(3/2))

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Maple [B]  time = 0.129, size = 644, normalized size = 5.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^3*(b*sec(f*x+e))^(3/2),x)

[Out]

1/8/f*(cos(f*x+e)+1)*(-1+cos(f*x+e))^3*(4*cos(f*x+e)^3*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)+8*cos(f*x+e)^2*(-c
os(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)+4*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)-16*cos(f*x+e)^2*(-cos(f*x+
e)/(cos(f*x+e)+1)^2)^(1/2)-5*cos(f*x+e)^2*arctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))-cos(f*x+e)^2*ln(-(2
*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^
(1/2)-1)/sin(f*x+e)^2)-4*cos(f*x+e)^2*ln(-2*(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+
2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)+5*cos(f*x+e)*arctan(1/2/(-cos(f*x+e)/(cos
(f*x+e)+1)^2)^(1/2))+cos(f*x+e)*ln(-(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*
x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)+4*cos(f*x+e)*ln(-2*(2*cos(f*x+e)^2*(-cos(f*x+e)/(
cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)+16*(-
cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))*(b/cos(f*x+e))^(3/2)*cos(f*x+e)^2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)/sin
(f*x+e)^8

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3*(b*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.89368, size = 991, normalized size = 8.77 \begin{align*} \left [\frac{10 \,{\left (b \cos \left (f x + e\right )^{2} - b\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} \sqrt{\frac{b}{\cos \left (f x + e\right )}}{\left (\cos \left (f x + e\right ) + 1\right )}}{2 \, b}\right ) + 5 \,{\left (b \cos \left (f x + e\right )^{2} - b\right )} \sqrt{-b} \log \left (\frac{b \cos \left (f x + e\right )^{2} + 4 \,{\left (\cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt{-b} \sqrt{\frac{b}{\cos \left (f x + e\right )}} - 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) + 8 \,{\left (5 \, b \cos \left (f x + e\right )^{2} - 4 \, b\right )} \sqrt{\frac{b}{\cos \left (f x + e\right )}}}{16 \,{\left (f \cos \left (f x + e\right )^{2} - f\right )}}, \frac{10 \,{\left (b \cos \left (f x + e\right )^{2} - b\right )} \sqrt{b} \arctan \left (\frac{\sqrt{\frac{b}{\cos \left (f x + e\right )}}{\left (\cos \left (f x + e\right ) - 1\right )}}{2 \, \sqrt{b}}\right ) + 5 \,{\left (b \cos \left (f x + e\right )^{2} - b\right )} \sqrt{b} \log \left (\frac{b \cos \left (f x + e\right )^{2} - 4 \,{\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt{b} \sqrt{\frac{b}{\cos \left (f x + e\right )}} + 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}\right ) + 8 \,{\left (5 \, b \cos \left (f x + e\right )^{2} - 4 \, b\right )} \sqrt{\frac{b}{\cos \left (f x + e\right )}}}{16 \,{\left (f \cos \left (f x + e\right )^{2} - f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3*(b*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

[1/16*(10*(b*cos(f*x + e)^2 - b)*sqrt(-b)*arctan(1/2*sqrt(-b)*sqrt(b/cos(f*x + e))*(cos(f*x + e) + 1)/b) + 5*(
b*cos(f*x + e)^2 - b)*sqrt(-b)*log((b*cos(f*x + e)^2 + 4*(cos(f*x + e)^2 - cos(f*x + e))*sqrt(-b)*sqrt(b/cos(f
*x + e)) - 6*b*cos(f*x + e) + b)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)) + 8*(5*b*cos(f*x + e)^2 - 4*b)*sqrt(b/
cos(f*x + e)))/(f*cos(f*x + e)^2 - f), 1/16*(10*(b*cos(f*x + e)^2 - b)*sqrt(b)*arctan(1/2*sqrt(b/cos(f*x + e))
*(cos(f*x + e) - 1)/sqrt(b)) + 5*(b*cos(f*x + e)^2 - b)*sqrt(b)*log((b*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 + co
s(f*x + e))*sqrt(b)*sqrt(b/cos(f*x + e)) + 6*b*cos(f*x + e) + b)/(cos(f*x + e)^2 - 2*cos(f*x + e) + 1)) + 8*(5
*b*cos(f*x + e)^2 - 4*b)*sqrt(b/cos(f*x + e)))/(f*cos(f*x + e)^2 - f)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**3*(b*sec(f*x+e))**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.40269, size = 181, normalized size = 1.6 \begin{align*} \frac{b^{6}{\left (\frac{5 \, \arctan \left (\frac{\sqrt{b \cos \left (f x + e\right )}}{\sqrt{-b}}\right )}{\sqrt{-b} b^{4}} + \frac{5 \, \arctan \left (\frac{\sqrt{b \cos \left (f x + e\right )}}{\sqrt{b}}\right )}{b^{\frac{9}{2}}} + \frac{2 \,{\left (5 \, b^{2} \cos \left (f x + e\right )^{2} - 4 \, b^{2}\right )}}{{\left (\sqrt{b \cos \left (f x + e\right )} b^{2} \cos \left (f x + e\right )^{2} - \sqrt{b \cos \left (f x + e\right )} b^{2}\right )} b^{4}}\right )} \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3*(b*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

1/4*b^6*(5*arctan(sqrt(b*cos(f*x + e))/sqrt(-b))/(sqrt(-b)*b^4) + 5*arctan(sqrt(b*cos(f*x + e))/sqrt(b))/b^(9/
2) + 2*(5*b^2*cos(f*x + e)^2 - 4*b^2)/((sqrt(b*cos(f*x + e))*b^2*cos(f*x + e)^2 - sqrt(b*cos(f*x + e))*b^2)*b^
4))*sgn(cos(f*x + e))/f